Aptitude Discussion

Q. |
To complete a piece of work $A$ and $B$ take 8 days, $B$ and $C$ 12 days. |

✖ A. |
7 Days |

✖ B. |
7.5 Days |

✔ C. |
8 Days |

✖ D. |
8.5 Days |

**Solution:**

Option(**C**) is correct

Given $(A+B)$'s one day’s work = \(\dfrac{1}{8}\)

$(B+C)$'s one day’s work = \(\dfrac{1}{12}\)

$(A+B+C) $'s 1 day’s work = \(\dfrac{1}{6}\)

Work done by $A$, alone:

$=(A+B+C)$'s 1 day’s work - $(B+C)$'s one day’s work

$= \dfrac{1}{6} - \dfrac{1}{12}$

$=\dfrac{1}{12}$

Work done by $C$, alone:

$=(A+B+C) $'s 1 day’s work** **- $(A+B)$'s one day’s work

$= \dfrac{1}{6} - \dfrac{1}{8}$

$=\dfrac{1}{24}$

$(A+C)$'s one day’s work:

$=\dfrac{1}{12}+\dfrac{1}{24}$

$=\dfrac{1}{8}$

$(A+C)$ will take $\textbf{8 days}$ to complete the work together.

**Edit:**For an alternative solution, check comment by **Manpreet Kunnath.**

**Abhijeet Panchalwar**

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Please explain how to solve further

By LCM method,

(A+B+C) (A+B) (B+C)

6 8 12 L.C.M=24

----------------------------

4 3 2

(A+B) +(B+C)+(C+A)= 2(A+B+C)= 2*4 units = 8 units

(A+B) +(B+C)+(C+A)= 8 units

(3 units)+(2 units)+ ((A+C) units)= 8 units

Therefore, A+C= 3 units

(A+B) (B+C) (A+C)

8 12 x L.C.M=24

----------------------------

3 2 3(i.e. units of A+C)

now, x=24/3= 8 days

**ABHI**

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A+B = 1/8th of the work

B+C = 1/12th of the work

A+B+C = 1/6th of the work

C= 1/6-1/8 =1/24

A= 1/6- 1/12 = 1/12

C+A = 1/24+ 1/12 = 3/24 i.e., 1/8th of the work

So togather they will take 8days

Hope you guys know the basics

**Pavan Kumar**

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Given (1/A+1/B)=1/8

(1/B+1/C)=1/12

(1/A+1/B+1/C)=1/6 Then

2* (1/A+1/B+1/C)=2/6 => (1/A+1/C)=2/6-1/8-1/12 =>

After solving above one we will get

(1/A+1/C)=1/8

so 8 days required to complete the work if A and C are worked to gether

How 1/A can be work done in one day ? Pls do reply

**Manpreet Kunnath**

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$A+B+C=6$ ..... (1)

$A+B=8$ ..... (2)

Replacing (2) in (1)

$8+C=6$

Hence, $C=2$ (Ignoring the negative symbol)

$B+C=12$ ..... (3)

replacing (3) in (1)

$A+12=6$

Hence, $A=6$ (Ignoring the negative symbol)

So adding the two results

$A+C = 6+2$

$= \textbf{8 days}$

How can you ignore negation?

Its the luck which helped u here.

**Suhail Ahmed**

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We have formula for this:

$=\dfrac{2XYZ}{XY+YZ+ZX}$

Simply solve for $=ZX$

**S Malik**

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@Shireen like $A+B=8$ SO $A=8-B$ and so on

Tried but could not work out the calculation, any help please.

**Muhammad Shehroz**

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Keep it simple. Put values of $A+B$ and $B+C$ in $A+B+C=6$ equation to get values of $C$ and $A$ simultaneously.

Can you please elaborate more?

Take LCM then check the option which divisible by 24 LCM value I, E 8days